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Math triangle fun please help!

Math triangle fun please help!-example-1
User Jfklein
by
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2 Answers

4 votes

Answer:

Explanation:

You can use the cosine rule to solve this problem:

cos(∠KLJ) = (KL² + JL² - KJ²)/(2*JL*KL) = 21463/21960 = 0.97737

∠KLJ = cos⁻¹(0.97737) = 12.2°

User Giliev
by
7.6k points
6 votes

Answer:

Explanation:

We would apply the law of Cosines which is expressed as

a² = b² + c² - 2abCosA

Where a,b and c are the length of each side of the triangle and A is the angle corresponding to a. Likening the expression to the given triangle, it becomes

JL² = JK² + KL² - 2(JK × KL)CosK

122² = 39² + 90² - 2(39 × 90)CosK

14884 = 1521 + 8100 - 2(3510)CosK

14884 = 9621 - 7020CosK

7020CosK = 9621 - 14884

7020CosK = - 5263

CosK = - 5263/7020

CosK = - 0.7497

K = Cos^- 1(- 0.7497)

K = 138.6° to the nearest tenth

User Rolfe
by
7.3k points

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