{"id":180561,"date":"2025-01-07T14:38:15","date_gmt":"2025-01-07T14:38:15","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=180561"},"modified":"2025-01-07T14:38:17","modified_gmt":"2025-01-07T14:38:17","slug":"1-6-meters-in-feet-rounded-to-the-nearest-tenth-using-the-following-conversion","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/07\/1-6-meters-in-feet-rounded-to-the-nearest-tenth-using-the-following-conversion\/","title":{"rendered":"1.6 meters in feet rounded to the nearest tenth using the following conversion"},"content":{"rendered":"\n

1.6 meters in feet rounded to the nearest tenth using the following conversion: 1 meter is 3.3 feet<\/p>\n\n\n\n

Solve for A to the nearest degree if 0\u00b0S AS 360\u00b0 13<\/p>\n\n\n\n

a) sin A<\/p>\n\n\n\n

b)cos A- – 1 V2 2 [K4]<\/p>\n\n\n\n

c) sec A = -2.5<\/p>\n\n\n\n

d) cot 4 = 3 [54]<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

1. Conversion from meters to feet:<\/h3>\n\n\n\n

To convert 1.6 meters to feet using the given conversion factor (1 meter = 3.3 feet):<\/p>\n\n\n\n

[
1.6 \\, \\text{meters} \\times 3.3 \\, \\text{feet\/meter} = 5.28 \\, \\text{feet}
]<\/p>\n\n\n\n

Rounded to the nearest tenth:<\/p>\n\n\n\n

[
5.28 \\, \\text{feet} \\approx 5.3 \\, \\text{feet}
]<\/p>\n\n\n\n

So, 1.6 meters is approximately 5.3 feet<\/strong>.<\/p>\n\n\n\n

2. Solve for A:<\/h3>\n\n\n\n

Let’s work through each part step-by-step.<\/p>\n\n\n\n

a) Solve for ( \\sin A ):<\/h4>\n\n\n\n

Given that we don’t have a specific triangle or further information, we can’t directly calculate ( \\sin A ) from the provided text. However, ( \\sin A ) typically relates to a triangle\u2019s opposite side over its hypotenuse in trigonometry. To solve for ( A ), we’d need more context, such as a right triangle’s dimensions or an angle.<\/p>\n\n\n\n

b) Solve for ( \\cos A ):<\/h4>\n\n\n\n

We are given:<\/p>\n\n\n\n

[
\\cos A = -\\frac{1}{\\sqrt{2}}
]<\/p>\n\n\n\n

This suggests that:<\/p>\n\n\n\n

[
\\cos A = -\\frac{1}{\\sqrt{2}} \\approx -0.7071
]<\/p>\n\n\n\n

Now, let\u2019s find the angle. The cosine function for ( -\\frac{1}{\\sqrt{2}} ) corresponds to an angle of:<\/p>\n\n\n\n

[
A = 135^\\circ \\, \\text{or} \\, A = 225^\\circ
]<\/p>\n\n\n\n

These values are based on the unit circle, where cosine equals ( -\\frac{1}{\\sqrt{2}} ) at these specific angles. Since the angle is between 0\u00b0 and 360\u00b0, both values are solutions, but the one most likely intended by the problem is 135\u00b0<\/strong>.<\/p>\n\n\n\n

c) Solve for ( \\sec A = -2.5 ):<\/h4>\n\n\n\n

We know that the secant function is the reciprocal of the cosine function:<\/p>\n\n\n\n

[
\\sec A = \\frac{1}{\\cos A}
]<\/p>\n\n\n\n

So, if ( \\sec A = -2.5 ), then:<\/p>\n\n\n\n

[
\\cos A = \\frac{1}{-2.5} = -0.4
]<\/p>\n\n\n\n

Now, we find the angle ( A ) corresponding to ( \\cos A = -0.4 ). Using a calculator:<\/p>\n\n\n\n

[
A = \\cos^{-1}(-0.4) \\approx 114.6^\\circ
]<\/p>\n\n\n\n

So, the angle ( A ) is approximately 114.6\u00b0<\/strong>.<\/p>\n\n\n\n

d) Solve for ( \\cot A = 3 ):<\/h4>\n\n\n\n

The cotangent function is the reciprocal of the tangent:<\/p>\n\n\n\n

[
\\cot A = \\frac{1}{\\tan A}
]<\/p>\n\n\n\n

Given ( \\cot A = 3 ), we can find ( \\tan A ):<\/p>\n\n\n\n

[
\\tan A = \\frac{1}{3} \\approx 0.3333
]<\/p>\n\n\n\n

Now, find the angle corresponding to ( \\tan A = 0.3333 ). Using the inverse tangent function:<\/p>\n\n\n\n

[
A = \\tan^{-1}(0.3333) \\approx 18.43^\\circ
]<\/p>\n\n\n\n

So, the angle ( A ) is approximately 18.4\u00b0<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Summary:<\/h3>\n\n\n\n
    \n
  • ( 1.6 \\, \\text{meters} ) is approximately 5.3 feet<\/strong>.<\/li>\n\n\n\n
  • For the trigonometric equations:<\/li>\n\n\n\n
  • ( \\sin A ) needs further information for calculation.<\/li>\n\n\n\n
  • ( \\cos A = -\\frac{1}{\\sqrt{2}} ) gives ( A = 135^\\circ ).<\/li>\n\n\n\n
  • ( \\sec A = -2.5 ) gives ( A \\approx 114.6^\\circ ).<\/li>\n\n\n\n
  • ( \\cot A = 3 ) gives ( A \\approx 18.4^\\circ ).<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

    1.6 meters in feet rounded to the nearest tenth using the following conversion: 1 meter is 3.3 feet Solve for A to the nearest degree if 0\u00b0S AS 360\u00b0 13 a) sin A b)cos A- – 1 V2 2 [K4] c) sec A = -2.5 d) cot 4 = 3 [54] The Correct Answer and […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-180561","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180561","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=180561"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180561\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=180561"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=180561"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=180561"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}